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2x + ½ = -5/3
2x = -5/3 - 1/2
2x = -13/6
x = -13/6 : 2
x = -13/12
1/7 - 3/5 x = 3/5
-3/5 x = 3/5 - 1/7
-3/5 x = 16/35
x = 16/35 : -3/5
x = -16/21
-3x - ¾ = 5/6
-3x = 5/6 + 3/4
-3x = 19/12
x = 19/12 : (-3)
x = -19/36
3/7 - 1/2 x = 5/3
-1/2 x = 5/3 - 3/7
-1/2 x = 26/21
x = 26/21 : (-1/2)
x = -52/21
2x - ¾ = 5/8
2x = 5/8 + 3/4
2x = 11/8
x = 11/8:2
x = 11/16
¼ x - |-7/5| = -5/3
¼ x - 7/5 = -5/3
¼ x = -5/7 + 7/5
¼ x = 24/35
x = 24/35 : 1/4
x = 96/35
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vì (2x-1)^6=(2x-1)^8
(2x-1)^8-(2x-1)^6=0
(2x-1)^6[(2x-1)^2-1)]=0
th1 (2x-1)^6 suy ra 2x-1=0 suy ra x=1/2
th2 (2x-1)^2-1=0
(2x-1)^2=1
suy ra 2x-1 bằng 1;-1
th1 2x-1=1 suy ra x=1
2x-1=-1 suy ra x=0
Bài làm:
a) \(2\left|x-1\right|-8=0\)
\(\Leftrightarrow2\left|x-1\right|=8\)
\(\Leftrightarrow\left|x-1\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
b) \(-\left|2x+3\right|+3=6\)
\(\Leftrightarrow\left|2x+3\right|=-3\)
Mà \(\left|2x+3\right|\ge0>-3\left(\forall x\right)\)
=> Mâu thuẫn
=> Không tồn tại x thỏa mãn
a) Ta có 2|x - 1| - 8 = 0
=> 2|x - 1| = 8
=> |x - 1| = 4
=> \(\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
b) Ta có : -|2x + 3| + 3 = 6
=> -|2x + 3| = 3
=> |2x + 3| = -3
Vì \(\left|2x+3\right|\ge0\forall x\)
mà -3 < 0
=> x \(\in\varnothing\)
\(a,\frac{8}{x}=\frac{x}{4}\)
\(=>x\cdot x=8\cdot4\)
\(=>x^2=32\)
\(=>x=\sqrt{32}\)
\(c,\frac{2x+3}{6}=\frac{x+1}{-8}\)
\(=>-8\cdot\left(2x+3\right)=6\cdot\left(x+1\right)\)
\(=>-16x-24=6x+6\)
\(=>-16x-6x=6+24\)
\(=>-22x=30\)
\(=>x=\frac{30}{-22}=-\frac{15}{11}\)
Bài giải
\(\left(2x-3\right)^6=\left(2x-3\right)^8\)
\(\left(2x-8\right)^8-\left(2x-3\right)^6=0\)
\(\left(2x-3\right)^6\left[\left(2x-3\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-3\right)^6=0\\\left(2x-3\right)^2-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)^2=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=3\\2x-3=-1\text{ hoặc }2x-3=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\text{ hoặc }x=2\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{3}{2}\text{ ; }1\text{ ; }2\right\}\)
\(\left(2x-3\right)^6=\left(2x-3\right)^8\)
\(\Rightarrow\left(2x-3\right)^8-\left(2x-3\right)^6=0\)
\(\left(2x-3\right)^6\left[\left(2x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-3\right)^6=0\\\left(2x-3\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)^2=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=1;x=2\end{cases}}}\)